P3128 [USACO15DEC]Max Flow P
考察:树上差分
思路:
点差分模板题.定义d[i] 为某路径上经过i的次数.
1 #include <iostream> 2 #include <cstring> 3 #include <queue> 4 using namespace std; 5 const int N = 50010; 6 int n,k,h[N],d[N],idx,depth[N],fa[N][16]; 7 struct Road{ 8 int fr,to,ne; 9 }road[N<<1]; 10 void add(int a,int b) 11 { 12 road[idx].fr = a,road[idx].to = b,road[idx].ne = h[a],h[a] = idx++; 13 } 14 void bfs(int s) 15 { 16 queue<int> q; 17 q.push(s); 18 memset(depth,0x3f,sizeof depth); 19 depth[s] = 1,depth[0] = 0; 20 while(q.size()) 21 { 22 int u = q.front(); 23 q.pop(); 24 for(int i=h[u];~i;i=road[i].ne) 25 { 26 int v = road[i].to; 27 if(depth[v]>depth[u]+1) 28 { 29 depth[v] = depth[u]+1; 30 q.push(v); 31 fa[v][0] = u; 32 for(int j=1;j<=15;j++) 33 fa[v][j] = fa[fa[v][j-1]][j-1]; 34 } 35 } 36 } 37 } 38 int lca(int a,int b) 39 { 40 if(depth[a]<depth[b]) swap(a,b); 41 for(int j=15;j>=0;j--) 42 if(depth[fa[a][j]]>=depth[b]) a = fa[a][j]; 43 if(a==b) return a; 44 for(int j=15;j>=0;j--) 45 if(fa[a][j]!=fa[b][j]) a = fa[a][j],b = fa[b][j]; 46 return fa[a][0]; 47 } 48 void dfs(int u,int fa) 49 { 50 for(int i=h[u];~i;i=road[i].ne) 51 { 52 int v = road[i].to; 53 if(v==fa) continue; 54 dfs(v,u); 55 d[u]+=d[v]; 56 } 57 } 58 int main() 59 { 60 scanf("%d%d",&n,&k); 61 for(int i=1;i<=n;i++) h[i] = -1; 62 for(int i=1;i<n;i++) 63 { 64 int a,b; 65 scanf("%d%d",&a,&b); 66 add(a,b); add(b,a); 67 } 68 bfs(1); 69 while(k--) 70 { 71 int a,b; scanf("%d%d",&a,&b); 72 int anc = lca(a,b); 73 d[a]++,d[b]++,d[anc]--,d[fa[anc][0]]--; 74 } 75 dfs(1,-1); 76 int ans = 0; 77 for(int i=1;i<=n;i++) ans = max(ans,d[i]); 78 printf("%d\n",ans); 79 return 0; 80 }
